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I have an equation as follows:

$$a \Delta \mathbf{u} + \mathbf{\nabla}(\mathbf{\nabla} \cdot \mathbf{u}) = 0$$

in which $a$ is a constant, $\mathbf{u}$ is a vector, $\Delta$ is the Laplacian operator, $\mathbf{\nabla}$ is the gradient operator and $\mathbf{\nabla} \cdot$ is the divergence operator.

If I take a Laplacian of the above equation, the answer is supposed to be:

$$\Delta \Delta \mathbf{u} = 0.$$

But how could I reach this?

(Actually, these two equations are eq.(7.4) and eq.(7.7) in the book of Theory of Elasticity - 1970 written by L. Landau)

$\bf{Edit.1}$ - A possible solution

Since we know: $\mathbf{\nabla}(\mathbf{\nabla} \cdot \mathbf{u}) = \Delta \mathbf{u} + \mathbf{\nabla} \times (\mathbf{\nabla} \times \mathbf{u})$, if we insert this equation to the first equation and then apply Laplacian operator on it, we will have a following equation:

$$ a \Delta \Delta \mathbf{u} + \Delta \Delta \mathbf{u} + \mathbf{\nabla} \cdot ( \mathbf{\nabla} ( \mathbf{\nabla} \times (\mathbf{\nabla} \times \mathbf{u}) ) ) = 0 $$

where I used the fact that $\Delta \mathbf{v} = \mathbf{\nabla} \cdot (\mathbf{\nabla} \mathbf{v} )$, in which $\mathbf{v}$ is another vector.

Now, if we assume that we can commute $\mathbf{\nabla}$ with $\mathbf{\nabla} \times $, the above equation would become:

$$ a \Delta \Delta \mathbf{u} + \Delta \Delta \mathbf{u} + \mathbf{\nabla} \cdot ( \mathbf{\nabla} \times ( \mathbf{\nabla} (\mathbf{\nabla} \times \mathbf{u}) ) ) = 0 $$

Since it is known that $\mathbf{\nabla} \cdot ( \mathbf{\nabla} \times \mathbf{v}) = 0$, finally we will have:

$$ (a + 1) \Delta \Delta \mathbf{u} = 0 $$

Then, either we say $a = -1$, or we say $\Delta \Delta \mathbf{u} = 0$.

It turns out that $a = 1- 2 \sigma$, in which $\sigma$ represents the Poisson's ratio. For a stable, isotropic and linear elastic material, $\sigma$ must be between -1.0 and 0.5. Thus, the possibility of $a = -1$ can be ruled out, and we have $\Delta \Delta \mathbf{u} = 0$ at the end.

Anyway, the possibility of the interchange of the gradient and curl would be critical for my approach.

$\bf{Edit.2}$ There is a problem in Edit.1

The curl of a gradient is 0, but it is not necessary that the gradient of a curl is also 0. Thus, the assumption that I made in Edit.1, which is that we can commute the curl and gradient, is wrong!

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  • $\begingroup$ Is there any statement made about the coordinate system? In cartesian coordinates, the Vector Laplacian boils down to $\boldsymbol \Delta \boldsymbol u = \Delta u_i$. Otherwise, I am not sure if you can prove this since you just bite always your own tail if you play around with the vector identities. $\endgroup$
    – Dan Doe
    Commented Apr 6, 2022 at 8:03
  • $\begingroup$ @DanDoe Since this is about a general statement of the property of displacements in the theory of elasticity, I think a 3D Cartesian coordinates is OK. $\endgroup$
    – ranger
    Commented Apr 6, 2022 at 11:56

1 Answer 1

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The problem can proved by this way.

If we take a divergence of the first equation, we get

$$ \Delta (\mathbf{\nabla} \cdot \mathbf{u}) = 0$$

Now, if we take the Laplacian of the first equation and take into account that a Laplacian and a gradient is commutable, we get

$$ a \Delta \Delta \mathbf{u} + \mathbf{\nabla}[\Delta (\mathbf{\nabla} \cdot \mathbf{u})] = 0 $$

then it is easy to see

$$\Delta \Delta \mathbf{u} = 0$$

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